
Biochemistry
9th Edition
ISBN: 9781319114671
Author: Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
Publisher: W. H. Freeman
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Question
![f 50
The Lineweaver-Burk plot below depicts the effect of an inhibitor on the initial velocity (Vo) of an
enzyme at different substrate concentrations [S] (with both values plotted as reciprocals and the line
extrapolated to negative values of [S]). Indicate whether the statements below are true or false:
1
akm
ulu/wrt of
1
VAX
max
6 - 31/1²
α=
Slope
1
(S) (MM)
[S]
α = 2
a=1
No inhibitor
=
(1). D
akm
Vmax](https://content.bartleby.com/qna-images/question/b1729e29-896c-44e4-b404-d435df4bfcb3/2bbdf49a-a061-4c5d-9151-7d19b929d4f0/7kg01j_thumbnail.jpeg)
Transcribed Image Text:f 50
The Lineweaver-Burk plot below depicts the effect of an inhibitor on the initial velocity (Vo) of an
enzyme at different substrate concentrations [S] (with both values plotted as reciprocals and the line
extrapolated to negative values of [S]). Indicate whether the statements below are true or false:
1
akm
ulu/wrt of
1
VAX
max
6 - 31/1²
α=
Slope
1
(S) (MM)
[S]
α = 2
a=1
No inhibitor
=
(1). D
akm
Vmax
![1
akm
Based on this diagram:
15
Vmax
No inhibitor
Slope =
[S] mm
The inhibitor increases the initial velocity of the enzyme
The inhibitor is not affecting the Vmax of the enzyme
HOC
JORD
The inhibitor is acting competitively
An increase in the inhibitor concentration results in a decrease in the apparent
Km of the enzyme
akm
Vmax
Choose...
Choose...
Choose... +
Choose... #
O
Report question issue
Not](https://content.bartleby.com/qna-images/question/b1729e29-896c-44e4-b404-d435df4bfcb3/2bbdf49a-a061-4c5d-9151-7d19b929d4f0/g9rdax4_thumbnail.jpeg)
Transcribed Image Text:1
akm
Based on this diagram:
15
Vmax
No inhibitor
Slope =
[S] mm
The inhibitor increases the initial velocity of the enzyme
The inhibitor is not affecting the Vmax of the enzyme
HOC
JORD
The inhibitor is acting competitively
An increase in the inhibitor concentration results in a decrease in the apparent
Km of the enzyme
akm
Vmax
Choose...
Choose...
Choose... +
Choose... #
O
Report question issue
Not
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- The KM values for the reaction of chymotrypsin with two different substrates are given in the table below. Considering this information, which substrate has the lower apparent affinity for the enzyme? Which substrate is likely to give a lower value for Vmax? Substrate N-acetylvaline ethyl ester N-acetyltyrosine ethyl ester KM (M) 8.8 X 10-² 6.6 X 10-4 N-acetylvaline ethyl ester has the lower apparent affinity for the enzyme; it will also likely to give a lower Vmax: N-acetyltyrosine ethyl ester has the lower apparent affinity for the enzyme; it will also likely to give the lower V₁ max. N-acetylvaline ethyl ester has the lower apparent affinity for the enzyme; N- acetyltyrosine ethyl ester is likely to give the lower Vmax: N-acetyltyrosine ethyl ester has the lower apparent affinity for the enzyme; N- acetylvaline will likely to give the lower Vmax. None of the above statements are correct.arrow_forwarda particular enzyme catalyzes a single reactant S to a single product P, following michaelis-menten kinetics rp=(VmaxCs) / (Km + Cs) 1. A reaction with this enzyme is carried out at very low substrate concentrations. Draw and label a curve on the plot that describes the reaction kinetics under those conditions.arrow_forwardThe Michaelis-Menten rate equation for reversible mixed inhibition is written as Vo = Vmax [S] aKm + a' [S] where Vo is initial velocity, Vmax is maximum velocity, [S] is substrate concentration, a represents the effect of the inhibitor bound to free enzyme (E), a' represents the effect of the inhibitor bound to the enzyme-substrate complex (ES), and Km is the Michaelis constant that represents the [S] at which the reaction reaches/Vm Vmax 2α' Derive an expression for the effect of a reversible inhibitor on apparent Km from the previous equation. Use the alphabet tab to enter a and the basic tab to enter the prime sign in your answer. = Apparent, or observed, Km is equivalent to the [S] at which Vo max. apparent Km =arrow_forward
- Some enzymes have catalytic activity only limited by diffusion. Which rate constants of an enzyme- catalyzed reaction is/are rate limiting for the enzyme? How does this line up/compare to the rate limiting step of Michaelis-Menten Enzyme Kinetics? (Please show work and correct answer)arrow_forwardWhat is the relative inhibition of an enzyme by a competitive inhibitor at [S] = KS and [I] = KI?arrow_forwardFrom a series of flasks with a constant concentration of enzyme the following initial velocities weretaken, they were obtained as a function of the concentration of the substrate.a) Calculate the KM and Vmax kinetic parameters of the three forms (Lineweaver-Burk, Eadie-Hofstee, Dixon).b) Analyze which are the atypical data that cause a low correlation, which can be eliminated and explain youranswer.arrow_forward
- NOTE: the enzyme-inhibitor complex requires 450 joule/mol to dissociateJOULE/MOLarrow_forwardLisa decides to obtain values for the Km and Vmax of an enzyme that was isolated from liver cells.. Using the Michaelis Menten plot. In what kind of measurements are needed and what would be needed to plot on a graph to estimate Km and Vmax?arrow_forwarda) Based on the data shown in the image, what are the Km and Vmax for the enzyme with L-DOPA and D-DOPA? Show any relevant analyses or calculatins you did to determine these values. ( HINT a graph might be helpful here! ) b) Based on your answer to part a, briefly describe how the kinetics of the enzyme differs for the two substrates. Which Substrate has better binding affinity to the enzymearrow_forward
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